- 64 Views
- Uploaded on
- Presentation posted in: General

Gaussian Elimination

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Gaussian Elimination

Chemical Engineering Majors

Author(s): Autar Kaw

http://numericalmethods.eng.usf.edu

Transforming Numerical Methods Education for STEM Undergraduates

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution

The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix

A set of n equations and n unknowns

. .

. .

. .

(n-1) steps of forward elimination

Step 1

For Equation 2, divide Equation 1 by and multiply by .

Subtract the result from Equation 2.

−

_________________________________________________

or

Repeat this procedure for the remaining equations to reduce the set of equations as

. ..

. ..

. ..

End of Step 1

Step 2

Repeat the same procedure for the 3rd term of Equation 3.

. .

. .

. .

End of Step 2

At the end of (n-1) Forward Elimination steps, the system of equations will look like

. .

. .

. .

End of Step (n-1)

Solve each equation starting from the last equation

Example of a system of 3 equations

. .

. .

. .

Start with the last equation because it has only one unknown

THE END

http://numericalmethods.eng.usf.edu

A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below:

Assuming g is the amount of Ni in organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates g is given by

The solution for the unknowns x1, x2, and x3 is given by

Find the values of x1, x2, and x3 using Naïve Gauss Elimination. Estimate the amount of nickel in organic phase when 2.3 g/l is in the aqueous phase using quadratic interpolation.

Number of steps of forward elimination is

(n-1)=(3-1)=2

Solution

Forward Elimination: Step 1

Yields

Forward Elimination: Step 1

Yields

Forward Elimination: Step 2

Yields

This is now ready for Back Substitution

Back Substitution: Solve for x3 using the third equation

Back Substitution: Solve for x2 using the second equation

Back Substitution:Solve for x1 using the first equation

The solution vector is

The polynomial that passes through the three data points is then

Where g is the amount of nickel in the organic phase and a is the amount of in the aqueous phase.

When 2.3 g/l is in the aqueous phase, using quadratic interpolation, the estimated amount of nickel in the organic phase is

THE END

http://numericalmethods.eng.usf.edu

Pitfall#1. Division by zero

Division by zero is a possibility at any step of forward elimination

Pitfall#2. Large Round-off Errors

Exact Solution

Pitfall#2. Large Round-off Errors

Solve it on a computer using 6 significant digits with chopping

Pitfall#2. Large Round-off Errors

Solve it on a computer using 5 significant digits with chopping

Is there a way to reduce the round off error?

- Increase the number of significant digits
- Decreases round-off error
- Does not avoid division by zero

- Gaussian Elimination with Partial Pivoting
- Avoids division by zero
- Reduces round off error

THE END

http://numericalmethods.eng.usf.edu

- Possible division by zero
- Large round-off errors

- Increase the number of significant digits
- Decreases round-off error
- Does not avoid division by zero

- Gaussian Elimination with Partial Pivoting
- Avoids division by zero
- Reduces round off error

At the beginning of the kth step of forward elimination, find the maximum of

If the maximum of the values is

in the p th row,

then switch rows p and k.

Which two rows would you switch?

Switched Rows

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution

Same as naïve Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination.

. .

. .

. .

THE END

http://numericalmethods.eng.usf.edu

Solve the following set of equations by Gaussian elimination with partial pivoting

- Forward Elimination
- Back Substitution

Number of steps of forward elimination is (n-1)=(3-1)=2

Examine absolute values of first column, first row

and below.

- Largest absolute value is 144 and exists in row 3.
- Switch row 1 and row 3.

Divide Equation 1 by 144 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2

Divide Equation 1 by 144 and

multiply it by 25, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Examine absolute values of second column, second row

and below.

- Largest absolute value is 2.917 and exists in row 3.
- Switch row 2 and row 3.

Divide Equation 2 by 2.917 and

multiply it by 2.667,

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Solving for a3

Solving for a2

Solving for a1

Consider the system of equations

In matrix form

=

Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping

Forward Elimination: Step 1

Examining the values of the first column

|10|, |-3|, and |5| or 10, 3, and 5

The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1.

Performing Forward Elimination

Forward Elimination: Step 2

Examining the values of the first column

|-0.001| and |2.5| or 0.0001 and 2.5

The largest absolute value is 2.5, so row 2 is switched with row 3

Performing the row swap

Forward Elimination: Step 2

Performing the Forward Elimination results in:

Back Substitution

Solving the equations through back substitution

Compare the calculated and exact solution

The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting

THE END

http://numericalmethods.eng.usf.edu

If a multiple of one row of [A]nxn is added or subtracted to another row of [A]nxn to result in [B]nxn then det(A)=det(B)

The determinant of an upper triangular matrix [A]nxn is given by

Using forward elimination to transform [A]nxn to an upper triangular matrix, [U]nxn.

Using naïve Gaussian elimination find the determinant of the following square matrix.

Divide Equation 1 by 25 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2

Divide Equation 1 by 25 and

multiply it by 144, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Divide Equation 2 by −4.8

and multiply it by −16.8,

.

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

After forward elimination

.

- Forward Elimination
- Back Substitution
- Pitfalls
- Improvements
- Partial Pivoting
- Determinant of a Matrix

For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html

THE END

http://numericalmethods.eng.usf.edu